Chapter 16 - Section 16.4 - Green''s Theorem - 16.4 Exercise - Page 1160: 7

$- \frac{9}{5}$

$$\eqalign{ & {\text{We have }}\int\limits_C {{x^2}{y^2}dx + y{{\tan }^{ - 1}}ydy} \cr & \cr & {\text{Identifying }}P\left( {x,y} \right)dx{\text{ and }}Q\left( {x,y} \right) \cr & \underbrace {\int\limits_C {{x^2}{y^2}dx + y{{\tan }^{ - 1}}ydy} }_{\int\limits_C {P\left( {x,y} \right)dx} + Q\left( {x,y} \right)dy} \cr & P\left( {x,y} \right) = {x^2}{y^2},{\text{ }}Q\left( {x,y} \right) = y{\tan ^{ - 1}}y \cr & \cr & {\text{Calculate the partial derivatives }}\frac{{\partial Q}}{{\partial x}}{\text{ and }}\frac{{\partial P}}{{\partial y}} \cr & \frac{{\partial Q}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {y{{\tan }^{ - 1}}y} \right] = 0 \cr & \frac{{\partial P}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {{x^2}{y^2}} \right] = 2{x^2}y \cr & \cr & {\text{Applying the Green's Theorem}} \cr & \int\limits_C {P\left( {x,y} \right)dx} + Q\left( {x,y} \right)dy = \iint\limits_D {\left( {\frac{{\partial Q}}{{\partial x}} - \frac{{\partial P}}{{\partial y}}} \right)}dA \cr & {\text{From the image shown below}}{\text{, we have the following region }}D \cr & D = \left\{ {\left. {\left( {x,y} \right) \in R} \right|0 \leqslant x \leqslant 1,0 \leqslant y \leqslant 3x} \right\} \cr & {\text{Therefore}} \cr & \int\limits_C {{x^2}{y^2}dx + y{{\tan }^{ - 1}}ydy} = \int_0^1 {\int_0^{3x} {\left( {0 - 2{x^2}y} \right)} } dydx \cr & = - 2\int_0^1 {\int_0^{3x} {{x^2}y} } dydx \cr & {\text{Integrating with respect to }}y \cr & = - 2\int_0^1 {\left[ {\frac{{{x^2}{y^2}}}{2}} \right]_0^{3x}} dx \cr & = - \int_0^1 {\left[ {{x^2}\left( {9{x^2}} \right) - 0} \right]} dx \cr & = - 9\int_0^1 {{x^4}} dx \cr & {\text{Integrating with respect to }}x \cr & = - 9\left[ {\frac{{{x^5}}}{5}} \right]_0^1 \cr & = - 9\left[ {\frac{{{{\left( 1 \right)}^5}}}{5} - \frac{{{{\left( 0 \right)}^5}}}{5}} \right] \cr & = - \frac{9}{5} \cr} $$